等腰三角形(easy) 用 python O(n3) 超时

这是为啥？

n = int(input())
points = []
for i in range(n):
    x, y = map(int, input().split())
    points.append((x, y))

count = 0
def calc(i, j):
    return (points[i][0]-points[j][0])**2 + (points[i][1]-points[j][1])**2

for i in range(n):
    for j in range(i+1, n):
        distij = calc(i, j)
        for k in range(j+1, n):
            # 判断第三个点不和前面两个点共线
            if (points[j][1] - points[i][1]) * (points[k][0] - points[i][0]) !=\
                    (points[k][1] - points[i][1]) * (points[j][0] - points[i][0]):
                distjk = calc(j, k)
                distik = calc(i, k)
                if distij == distik or distij == distjk or distik == distjk:
                    count += 1

print(count)